[SOLVED] Help finishing my proof for: Prove that inf A = -sup(-A)

Let A be a nonempty subset of R (the reals) and let -A = {-x : x in A}.
Prove that inf A = -sup (-A).

My work:

Proof: Let A be non empty subset of R and -A = {-x : x in A}. If -x is in -A then
-x < sup (-A) by the definition of supremum. This implies x > -sup (-A), and so
-sup (-A) is a lower bound of A.

Now that I look back on my work so far, I think I am wrong in assuming right away that a supremum exists. Do I have to first prove the existence of a supremum than show it is equal to the infimum? I am really stuck here.
Any help is greatly appreciated. Thank you for your time!

Let A be a nonempty subset of R (the reals) and let -A = {-x : x in A}.
Prove that inf A = -sup (-A).

My work:

Proof: Let A be non empty subset of R and -A = {-x : x in A}. If -x is in -A then
-x < sup (-A) by the definition of supremum. This implies x > -sup (-A), and so
-sup (-A) is a lower bound of A.

your work is incomplete. showing it is a lower bound is not enough. you need to show it is the greatest lower bound. there are ways to do that elegantly. i will use the fact that for .

Let and be as defined. By the definition of infimum, we have that for all . so that for all . but that means is an upper bound for . thus, since the supremum is the least upper bound, we must have ........(1)

Also, for all by the definition of supremum. but that means for all , so that is a lower bound for the set . since the infimum is the greatest lower bound, we must have that ...........(2)

By (1) and (2) we have , as desired

Now that I look back on my work so far, I think I am wrong in assuming right away that a supremum exists. Do I have to first prove the existence of a supremum than show it is equal to the infimum? I am really stuck here.
Any help is greatly appreciated. Thank you for your time!

yes, we can assume the supremum and infimum exist. they are and in the extreme cases